If , prove that : 3 cos B – 4cos³ B = 0

Given:

We know that,

Or

Let,

Perpendicular =AB =k

and Hypotenuse =AC =2k

where, k is any positive integer

So, by Pythagoras theorem, we can find the third side of a triangle

In right angled ∆ ABC, we have

⇒ (AB)² + (BC)² = (AC)²

⇒ (k)² + (BC)² = (2k)²

⇒ k² + (BC)² = 4k²

⇒ (BC)² = 4k² –k²

⇒ (BC)² = 3k²

⇒ BC =√3k²

⇒ BC =k√3

So, BC = k√3

Now, we have to find the value of cos B

We know that,

Side adjacent to angle B or base = BC = k√3

Hypotenuse = AC =2k

So,

Now, LHS = 3 cos B – 4cos³ B

=RHS

Hence Proved